Step 1: Set up the two half-reactions with electrons balanced

$\ce{PbO2(s) + 4H+(aq) + 4e- -> Pb(s) + 2H2O(l)}$, $E^{\circ}_{1}=0.671\text{ V}$

$\ce{PbO(s) + H2O(l) + 2e- -> Pb(s) + 2OH-(aq)}$, $E^{\circ}_{2}=-0.580\text{ V}$

We need $\ce{2PbO(s) -> PbO2(s) + Pb(s)}$, i.e. one PbO is oxidized to PbO2 and one PbO is reduced to Pb. Reverse half-reaction 1 (oxidation) and double half-reaction 2 (reduction) so electrons cancel (4 each):

• Oxidation: $\ce{Pb(s) + 2H2O(l) -> PbO2(s) + 4H+(aq) + 4e-}$, $\Delta G_{A} = +4F(0.671) = +258.98\text{ kJ}$
• Reduction (×2): $\ce{2PbO(s) + 2H2O(l) + 4e- -> 2Pb(s) + 4OH-(aq)}$, $\Delta G_{B} = 2\times[-2F(-0.580)] = +223.84\text{ kJ}$

Step 2: Add and simplify

Adding gives:

$\ce{2PbO(s) + Pb(s) + 4H2O(l) -> PbO2(s) + 2Pb(s) + 4H+(aq) + 4OH-(aq)}$

Cancel one Pb and note the $\ce{4H+}$ and $\ce{4OH-}$ must combine back to $\ce{4H2O}$ to give the target reaction ($\ce{2PbO(s) -> PbO2(s) + Pb(s)}$). This neutralization step is not free — it has $\Delta G^{\circ} = RT\ln K_w$ per mole of $\ce{H2O}$ formed.

$\Delta G^{\circ}_{\text{neutralization, 1 mol}} = RT\ln(10^{-14}) = (8.314)(298)(-32.24) = -79.9\text{ kJ}$

Since 4 mol of water are formed: $\Delta G_{\text{neut}} = 4(-79.9) = -319.6\text{ kJ}$

Step 3: Total

$\Delta G^{\circ} = \Delta G_{A} + \Delta G_{B} + \Delta G_{\text{neut}} = 258.98 + 223.84 - 319.6 \approx 163\text{ kJ mol}^{-1}$

This matches choice (A).